homeomorphism

i thought if is homeomorphism, then is and on but not nessesarily entire and the inverse exists on . so i thought the question emphasized that . i dont know if that is right but that is what i thought.

i thought if is homeomorphism, then is and on but not nessesarily entire and the inverse exists on . so i thought the question emphasized that . i dont know if that is right but that is what i thought.

Well typically if the function is defined to be and is onto, it is usually assumed to be onto the given range in my experience. Emphasis never hurts though

I think I have it. Suppose . Then is not closed (since only closed and open sets are the empty and whole sets). In particular there exists a such that there exists a sequence with each . Then since the sequence is convergent, it is Cauchy, and since the function is uniformly continuous the sequence of images is Cauchy. Since is complete, there exists such that . Since the function was onto, there exists u such that .

Then since is continuous, implies . But so by uniqueness of limits . Since we assumed but , we have our contradiction!

how do you know that the convergent sequence is Cauchy? i know that every Cauchy sequence converges in but does the converse holds too?

Yes! The proof is easy too. Suppose . Let be given. Then by the definition of convergence there exists a such that for all , . Then for all , .

Convergence is a stronger condition than Cauchy since to be convergent we must identify the element the sequence converges to, whereas with cauchy we only know the sequence elements get closer together.